Usaco 2.24 party lamps

Initialization. All lights are on,

Then there are only four operations in total

1) All lights are reversed.

2) invert of the lamp numbered with an odd number

3) invert the number of lights with an even number

4) invert the light numbered 3 * x + 1

Now, ask, after performing Step C, some lights are on, some lights are off, and the possible combination of all the lights is obtained.

Question:

The effect of pressing a button twice is the same as that of a button without a button. Therefore, each button or press or do not press, a total of 2 ^ 4 = 16 states. Enumerate whether or not each button is pressed, generate the result, and sort the output (pay attention to the weight ).

/* ID: nanke691lang: C ++ task: lamps */# include <iostream> # include <fstream> # include <string> # include <algorithm> # include <set> using namespace STD; int N, C, T1, t2; int on [110] = {0}, off [110] = {0}; set <string> S; void judge (int A, int B, int C, int d) {string STR = ""; for (INT I = 1; I <= N; I ++) {int T = A; If (I-1) % 3 = 0) T + = D; if (I % 2 = 0) T + = C; else t + = B; If (T % 2 = 0) STR + = "1"; else STR + = "0";} For (INT I = 0; I <t1; I ++) {// cout <"on" <''<on [I] <<Endl; If (STR [ON [I]-1] = '0') return ;}for (INT I = 0; I <t2; I ++) {// cout <"off" <''<off [I] <Endl; If (STR [off [I]-1] = '1 ') return;} s. insert (STR);} int main () {freopen ("lamps. in "," r ", stdin); freopen (" lamps. out "," W ", stdout); int A; CIN> N> C; CIN> A; T1 = 0, T2 = 0; while (! =-1) {ON [T1 ++] = A; CIN> A;} CIN> A; while (! =-1) {off [T2 ++] = A; CIN> A ;}for (INT I = 0; I <2; I ++) for (Int J = 0; j <2; j ++) for (int K = 0; k <2; k ++) for (int l = 0; L <2; l ++) {int c = I + J + K + L; If (C % 2 = C % 2 & C <= C) judge (I, J, K, L);} set <string >:: iterator it = S. begin (); If (S. size ()! = 0) {for (; it! = S. End (); It ++) cout <* It <Endl;} else cout <"impossible" <Endl ;}